Standard Proofs by Induction (Edexcel IAL Further Maths)

Proof by Induction with Series

Spec point — 8.1

What are the steps for proof by induction with series?

• For example: prove that $\sum_{r = 1}^{n} r^{2} = \frac{1}{6} n ( n + 1 ) ( 2 n + 1 )$* *

  • It has a left-hand side, $L H S = \sum_{r = 1}^{n} r^{2}$
  • and a right-hand side, $RHS=\frac{1}{6}n(n+1)(2n+1)$

• STEP 1 The basic step: Show result is true for $n=1$

  • Substitute *$n=1$ into both sides individually $L H S = \sum_{r = 1}^{1} r^{2} = 1^{2} = 1 R H S = \frac{1}{6} \times 1 \times ( 1 + 1 ) ( 2 \times 1 + 1 ) = \frac{1}{6} \times 2 \times 3 = 1$

• STEP 2 The assumption step: Assume the LHS and RHS are equal for some $n=k$* where $k$ *is some integer

  • Replace $n$ with $k$ in the statement
  • Write: "Assume $\sum_{r = 1}^{k} r^{2} = \frac{1}{6} k ( k + 1 ) ( 2 k + 1 )$ is true"

• STEP 3 The inductive step: You need to prove that the LHS and RHS are equal for $n=k+1$

  • Start with the LHS * for $n=k+1$

    • Take the last term out: $\sum_{r = 1}^{k + 1} r^{2 } = \sum_{r = 1}^{k} r^{2} + ( k + 1 )^{2}$
    • Substitute in the assumption(STEP 2) :
    • $\sum_{r = 1}^{k + 1} r^{2 } = \frac{1}{6} k ( k + 1 ) ( 2 k + 1 ) + ( k + 1 )^{2}$
    • Factorise: $\frac{1}{6} ( k + 1 ) [ k ( 2 k + 1 ) + 6 ( k + 1 ) ] = \frac{1}{6} ( k + 1 ) [ 2 k ^{2} + 7 k + 6 ] = \frac{1}{6} ( k + 1 ) ( k + 2 ) ( 2 k + 3 )$

  • Check if it is the same as the RHS for $n=k+1$ $\begin{gathered} R H S & = \frac{1}{6} ( k + 1 ) ( ( k + 1 ) + 1 ) ( 2 ( k + 1 ) + 1 ) \\ = \frac{1}{6} ( k + 1 ) ( k + 2 ) ( 2 k + 3 ) \end{gathered}$

  • So LHS = RHS for $n=k+1$, as long as the assumption is true

• STEP 4 The conclusion step: Explain in words how the above steps make the result true for all integers, using the following two sentences:

  • "If it is true for $n=k$,* then it is true for $n=k+1$."
  • "As it is true for $n=1$, the statement is true for all $n\inℤ^{+}$."

Worked Example Prove by induction that $\sum_{r = 1}^{n} r ( r - 3 ) = 1 \frac{1}{3} n ( n - 4 ) ( n + 1 )$ for $n\inℤ^{+}$.

STE P 1: The basic step Find the LHS when $n=1$

$\sum_{r = 1}^{1} r ( r - 3 ) = 1 ( 1 - 3 ) = - 2$* *

Find the RHS when* *$n=1$

$1 \frac{1}{3} \times 1 \times ( 1 - 4 ) ( 1 + 1 ) = 1 \frac{1}{3} \times ( - 6 ) = - 2$* *

The LHS and RHS are equal, which shows the statement is true for $n=1$ State this

LHS = RHS, so it is true for $n=1$

STEP 2: The assumption step Assume the statement is true for some $n=k$ where $k$* *is a positive integer

Assume that $\sum_{r = 1}^{k} r ( r - 3 ) = 1 \frac{1}{3} k ( k - 4 ) ( k + 1 )$

STEP 3: The inductive step Find the RHS when $n=k+1$

$1 \frac{1}{3} ( k + 1 ) ( k + 1 - 4 ) ( k + 1 + 1 ) = 1 \frac{1}{3} ( k + 1 ) ( k - 3 ) ( k + 2 )$* *

Find the LHS when $n=k+1$

$\sum_{r = 1}^{k + 1} r ( r - 3 )$

Prove that the LHS when $n=k+1$ equals the RHS when $n=k+1$ Start by pulling out the final term in the sum

$\begin{gathered} \sum_{r = 1}^{k + 1} r ( r - 3 ) & = \sum_{r = 1}^{k} r ( r - 3 ) + ( k + 1 ) ( k + 1 - 3 ) \\ = \sum_{r = 1}^{k} r ( r - 3 ) + ( k + 1 ) ( k - 2 ) \end{gathered}$

Now use the assumption in STEP 2 to replace the sum from 1 to $k$ with its answer

$\begin{gathered} \sum_{r = 1}^{k + 1} r ( r - 3 ) & = 1 \frac{1}{3} k ( k - 4 ) ( k + 1 ) + ( k + 1 ) ( k - 2 ) \end{gathered}$

Now factorise the right-hand side

$\begin{gathered} \sum_{r = 1}^{k + 1} r ( r - 3 ) & = ( k + 1 ) [ 1 \frac{1}{3} k ( k - 4 ) + ( k - 2 ) ] \\ = 1 \frac{1}{3} ( k + 1 ) [ k ( k - 4 ) + 3 ( k - 2 ) ] \end{gathered}$

Expand inside the square brackets then continue to factorise

$\begin{gathered} \sum_{r = 1}^{k + 1} r ( r - 3 ) & = 1 \frac{1}{3} ( k + 1 ) [ k^{2} - 4 k + 3 k - 6 ] \\ = 1 \frac{1}{3} ( k + 1 ) [ k^{2} - k - 6 ] \\ = 1 \frac{1}{3} ( k + 1 ) ( k - 3 ) ( k + 2 ) \end{gathered}$

This is the same as the RHS when $n=k+1$ The LHS and RHS are equal, which shows that the statement is true for $n=k+1$ State this

LHS = RHS, so it is true for $n=k+1$

STEP 4: The conclusion We have just proved that if the case when $n=k$ is true, then the case when $n=k+1$ is true We have also shown that the case when $n=1$ is true Those two parts come together to mean that $n=1$* is true, so** *$n=2$ is true, so $n=3$ is true, ... and so on

Final answer: If it is true for$n=k$, then it is true for$n=k+1$. As it is true for$n=1$, the statement is true for all$n\inℤ^{+}$.

Proof by Induction with Divisibility

Spec point — 8.1

What are the steps for proof by induction with divisibility?

• For example: prove that $f(n)=4^{n}-1$ is divisible by 3 for all positive integers n

  • Being divisible by 3 is the same as being a multiple of 3

• STEP 1 The basic step: Show result is true for $n=1$

  • Substitute *$n=1$ into the function
  • $\mathrm{f} ( 1 ) = 4^{1} - 1 = 3$which is divisible by 3

• STEP 2 The assumption step: Assume $f(k)=4^{k}-1$ is divisible by 3 for some $n=k$* where $k$ *is some integer

  • Replace $n$ with $k$ in the statement
  • Write "divisible by 3" as "$=3m$" where $m$ is a positive integer
  • So$\mathrm{f} ( k ) = 4^{k} - 1 = 3 m$ where $m\inℤ^{+}$
  • It helps to make $4^{k}$ the subject: $4^{k}=1+3m$

• STEP 3 The inductive step (method 1): You need to show that the $n=k+1$ case is divisible by 3

  • Substitute $n=k+1$ into the function

    • $f(k+1)=4^{k+1}-1$

  • Use index laws to substitute in the assumption(STEP 2) $4^{k}=1+3m$

    • $\mathrm{f} ( k + 1 ) = 4 ( 4^{k} ) - 1 = 4 ( 1 + 3 m ) - 1 = 3 ( 4 m + 1 )$which is a multiple of 3

  • So $f(k+1)$is divisible by 3 (as long as the assumption is true)

• STEP 3 The inductive step (method 2): An alternative method is to show that the difference $f(k+1)-f(k)$ is a multiple of 3

  • i.e. $f(k+1)-f(k)=3\times...$

  • Making $f(k+1)$the subject gives $\mathrm{f} ( k + 1 ) = \mathrm{f} ( k ) + 3 \times...$

  • So $f(k+1)$ is divisible by 3 (as long as the assumption is true, i.e. that $f(k)$ is divisible by 3)

  • This method does not always work

    • When it does, a hint will be given in the question

• STEP 4 The conclusion step: Explain in words how the above steps make the result true for all integers, using the following two sentences:

  • "If it is true for $n=k$,* then it is true for $n=k+1$."
  • "As it is true for $n=1$, the statement is true for all $n\inℤ^{+}$."

Worked Example Prove by induction that $f(n)=6^{n}+13^{n+1}$ is divisible by 7 for all integers satisfying $n\geq0$.

STE P 1: The basic step Prove that it is true for $n=0$

$\begin{gathered} \mathrm{f} ( 0 ) & = 6^{0 + 13^{0} + 1} \\ = 1 + 13 \\ = 14 \\ = 7 \times 2 \end{gathered}$

State that it is true for $n=0$

14 is divisible by 7, so it is true for $n=0$

STEP 2: The assumption step Assume the statement is true for some $n=k$ where $k$ is a positive integer

Assume that $f(k)=6^{k}+13^{k+1}$ is divisible by 7

Replace "being divisible by 7" with being equal to $7m$, where $m$ is a positive integer

$6^{k}+13^{k+1}=7m$

STEP 3: The inductive step Substitute $n=k+1$ into the function

$\begin{gathered} \mathrm{f} ( k + 1 ) & = 6^{k + 1} + 13^{( k + 1 ) + 1} \\ = 6^{k + 1} + 13^{k + 2} \end{gathered}$

Now make either $6^{k}$ or $13^{k+1}$ the subject of STEP 2 and substitute it in It helps to use index laws first

$\begin{gathered} \mathrm{f} ( k + 1 ) & = 6^{k \times 6 + 13^{k} + 2} \\ = ( 7 m - 13^{k + 1} ) \times 6 + 13^{k + 2} \end{gathered}$

Simplify the terms, using index laws to group the $13^{k+1}$ terms

$\begin{gathered} \mathrm{f} ( k + 1 ) & = 42 m - 6 \times 13^{k + 1} + 13^{k + 2} \\ = 42 m - 6 \times 13^{k + 1} + 13 \times 13^{k + 1} \\ = 42 m + ( - 6 + 13 ) \times 13^{k + 1} \\ = 42 m + 7 \times 13^{k + 1} \end{gathered}$

The goal is to show that this result is divisible by 7 Factorise out a 7 to show this

$\begin{gathered} \mathrm{f} ( k + 1 ) & = 7 \times ( 6 m + 13^{k + 1} ) \end{gathered}$

State that it is true for $n=k+1$

$\mathrm{f} ( k + 1 )$ is divisible by 7, so it is true for $n=k+1$

STEP 4: The conclusion We have just proved that if the case when $n=k$ is true, then the case when $n=k+1$ is true We have also shown that the case when $n=0$ is true Those two parts come together to mean that $n=0$* is true, so** *$n=1$ is true, so $n=2$ is true, ... and so on

Final answer: If it is true for$n=k$, then it is true for$n=k+1$. As it is true for$n=0$, the statement is true for all integers$n\geq0$.

Proof by Induction with Sequences

Spec point — 8.1

What are the steps for proof by induction with sequences?

• For example: prove that the sequence given recursively by $u_{n+1}=3u_{n}+4$ where $u_{1}=1$ has the n^th term formula $u_{n}=3^{n}-2$ for $n\geq1$

• STEP 1 The basic step: Show result is true for $n=1$

  • Substitute *$n=1$ into the formulae separately
  • $u_{1}=1$ from the recursive formula
  • $u_{1}=3^{1}-2=1$ from the n^th term formula

• STEP 2 The assumption step: Assume that the term $u_{k}$ satisfies* $u_{k}=3^{k}-2$ for some $n=k$ where $k$ *is some integer

  • Replace $n$ with $k$ in the statement

  • The assumption is on the n^th term formula

    • (not on the recursive formula)

• STEP 3 The inductive step: Show that the n^th term formula is true for $n=k+1$

  • Use the recursive formula to write $u_{k+1}=3u_{k}+4$

  • Substitute in the assumption (STEP 2) and simplify

    • $u_{k + 1} = 3 ( 3^{k - 2 ) + 4 = 3^{k} + 1} - 2$

  • Check this is the same as the* n*th term formula when $n=k+1$

    • $u_{k+1}=3^{k+1}-2$
    • It is the same

• STEP 4 The conclusion step: Explain in words how the above steps make the result true for all integers, using the following two sentences:

  • "If it is true for $n=k$,* then it is true for $n=k+1$."
  • "As it is true for $n=1$, the statement is true for all $n\inℤ^{+}$."

Exam Hint

• If the recursive formula involves the previous two terms, then the basic step must be done for both $n=1$ and $n=2$.

Worked Example A sequence is defined by $u_{n+1}=5u_{n}-4$ where $u_{1}=2$ and $n\geq1$.

Prove by mathematical induction that $u_{n}=5^{n-1}+1$, where $n\geq1$.

STE P 1: The basic step Find $u_{1}$ from the recursive sequence

$u_{1}=2$

Find $u_{1}$* from the *n^th term formula

$\begin{gathered} u_{1} & = 5^{1 - 1} + 1 \\ = 5^{0} + 1 \\ = 1 + 1 \\ = 2 \end{gathered}$

These two terms are equal State that it is true for $n=1$

It is true for $n=1$

STEP 2: The assumption step Assume the formula is true for some $n=k$ where $k$ is a positive integer

Assume that $u_{k}=5^{k-1}+1$

STEP 3: The inductive step Use the recursive sequence to write $u_{k+1}$ in terms of $u_{k}$

$\begin{gathered} u_{k + 1} & = 5 u_{k} - 4 \end{gathered}$

Replace $u_{k}$ with the assumption in STEP 2 and simplify

$\begin{gathered} u_{k + 1} & = 5 ( 5^{k - 1} + 1 ) - 4 \\ = 5^{k} + 5 - 4 \\ = 5^{k} + 1 \end{gathered}$

Check that this is the same as substituting $n=k+1$* into the *n^th term formula

$\begin{gathered} u_{k + 1} & = 5^{( k + 1 ) - 1} + 1 \\ = 5^{k} + 1 \end{gathered}$

State that the result is true for $n=k+1$

It is true for $n=k+1$

STEP 4: The conclusion We have just proved that if the case when $n=k$ is true, then the case when $n=k+1$ is true We have also shown that the case when $n=1$ is true Those two parts come together to mean that $n=1$* is true, so** *$n=2$ is true, so $n=3$ is true, ... and so on

Final answer: If it is true for$n=k$, then it is true for$n=k+1$. As it is true for$n=1$, the statement is true for all$n\inℤ^{+}$.

Proof by Induction with Matrices

Spec point — 8.1

What are the steps for proof by induction with matrices?

• For example: prove that $M^{n} = ( \begin{gathered} 2^{n} & 0 \\ 1 - 2^{n} & 1 \end{gathered} )$ where $M = ( \begin{gathered} 2 0 \\ - 1 & 1 \end{gathered} )$ for all integers $n\geq1$.

• STEP 1 The basic step: Show result is true for $n=1$

  • Substitute *$n=1$ into formula
  • $M^{1 =} ( \begin{gathered} 2^{1} & 0 \\ 1 - 2^{1} & 1 \end{gathered} ) = ( \begin{gathered} 2 0 \\ - 1 & 1 \end{gathered} ) = M$

• STEP 2 The assumption step: Assume $M^{k} = ( \begin{gathered} 2^{k} & 0 \\ 1 - 2^{k} & 1 \end{gathered} )$ is true for some $n=k$* where $k$ *is some integer

  • Replace $n$ with $k$ in the statement

• STEP 3 The inductive step: You need to show that the $n=k+1$ case is true

  • Use index laws to write $M^{k+1}=\mathrm{MM}^{k}$

  • Substitute in the assumption: $M^{k + 1} = M ( \begin{gathered} 2^{k} & 0 \\ 1 - 2^{k} & 1 \end{gathered} ) = ( \begin{gathered} 2 0 \\ - 1 & 1 \end{gathered} ) ( \begin{gathered} 2^{k} & 0 \\ 1 - 2^{k} & 1 \end{gathered} )$

  • Multiply the matrices

    • M^{k + 1} = ( \begin{gathered} 2^{k + 1} & 0 \\ 1 - 2 \times 2^{k & 1 \end{gathered} ) = ( \begin{gathered} 2^{k} + 1} & 0 \\ 1 - 2^{k + 1} & 1 \end{gathered} )

  • Check if it is the same as the formula with $n=k+1$

    • $M^{k + 1} = ( \begin{gathered} 2^{k + 1} & 0 \\ 1 - 2^{k + 1} & 1 \end{gathered} )$
    • It is the same

• STEP 4 The conclusion step: Explain in words how the above steps make the result true for all integers, using the following two sentences:

  • "If it is true for $n=k$,* then it is true for $n=k+1$."
  • "As it is true for $n=1$, the statement is true for all $n\inℤ^{+}$."

Worked Example The matrix $M$ is given by $M = ( \begin{gathered} 2 2 \\ 0 1 \end{gathered} )$.

Prove, using mathematical induction, that $M^{n} = ( \begin{gathered} 2^{n} & 2 ( 2^{n} - 1 ) \\ 0 1 \end{gathered} )$ for all integers satisfying $n\geq1$.

STE P 1: The basic step Prove that it is true for $n=1$

$\begin{gathered} M^{1} & = ( \begin{gathered} 2^{1} & 2 ( 2^{1} - 1 ) \\ 0 1 \end{gathered} ) \\ = ( \begin{gathered} 2 2 \\ 0 1 \end{gathered} ) \\ = M \end{gathered}$

State that it is true for $n=1$

It is true for $n=1$

STEP 2: The assumption step Assume the formula is true for some $n=k$ where $k$ is a positive integer

Assume that $M^{k} = ( \begin{gathered} 2^{k} & 2 ( 2^{k} - 1 ) \\ 0 1 \end{gathered} )$* *

STEP 3: The inductive step Look at the power of $n=k+1$

$\begin{gathered} M^{k + 1} \end{gathered}$

Use index laws, then substitute in the assumption from STEP 2

| M ^ k + 1 | = M ^ k M ^ 1 | | _ = ( | 2 ^ k | 2 ( 2 ^ k - 1 ) | | 0 1 | ) ( | 2 2 | 0 1 | ) | |

Use matrix multiplication to work out the right-hand side Simplify the terms

• *\begin{gathered} M^{k + 1} & = ( \begin{gathered} 2^{k \times 2 + 0 & 2^{k} \times 2 + 2 ( 2^{k} + 1} - 1 ) \times 1 \\ 0 + 0 & 0 + 1 \end{gathered} ) \\ = ( \begin{gathered} 2^{k + 1} & 2 \times 2^{k + 1} - 2 \\ 0 1 \end{gathered} ) \end{gathered}

Check that this is the same as substituting $n=k+1$ into the formula for $M^{n}$

• *$\begin{gathered} M^{k + 1} & = ( \begin{gathered} 2^{k + 1} & 2 ( 2^{k + 1} - 1 ) \\ 0 1 \end{gathered} ) \\ = ( \begin{gathered} 2^{k + 1} & 2 \times 2^{k + 1} - 2 \\ 0 1 \end{gathered} ) \end{gathered}$

State that the result is true for $n=k+1$

It is true for $n=k+1$

STEP 4: The conclusion We have just proved that if the case when $n=k$ is true, then the case when $n=k+1$ is true We have also shown that the case when $n=1$ is true Those two parts come together to mean that $n=1$* is true, so** *$n=2$ is true, so $n=3$ is true, ... and so on

Final answer: If it is true for$n=k$, then it is true for$n=k+1$. As it is true for$n=1$, the statement is true for all$n\inℤ^{+}$.